splice is affecting previously copied variables [duplicate]

Answer 1

In first case you are working with arrays, which are passed by reference. And in second case you are working with prime types which are passed by value. In first case you should copy initial array (e.g. with initial_variable.slice(0)). Try something like

window.onload = function() {
  var initial_variable = ['first', 'second', 'third'];
  var copy_initial_variable = initial_variable.slice(0); //returns new array!!!!
  copy_initial_variable.splice(0, 1);
  alert('initial variable - ' + initial_variable);
};

Answer 2

The problem isn't in how splice behaves, but the fact that initial_variable and copy_initial_variable reference the same array.

alert (copy_initial_variable === initial_variable);

In JavaScript there are two types of values: primitive values, like numbers and booleans, and objects, including arrays. Variables hold primitive values, but they hold references to objects. "Copying" primitive values works as you expect, a new primitive value is created so that changing the copy variable will not change the original variable. But "copying" an object actually copies the reference pointing to that object, it doesn't create a new object.

It is not a JavaScript bug, this is the intended behavior.

Answer 3

This is in no way a bug, but a very common misunderstanding. Let's see what happens when I say

var a = b;

Integers and other javascript primitives, like floats and booleans, are "assigned by value". Which means that whatever value b has is going to be copied to a. To the computer, it means having the part of memory that b references copied to the memory that a references. That's the behavior you were expecting.

When arrays and other objects (and "descendants" of a new Object() call) are used like that, there is a copy by reference. Meaning that the value of a now references the value of b, the memory that b references isn't copied or modified. Thus, when writing

a = [1,2,3];
b = a;

b and a become interchangeable. They're referencing the same memory address. To achieve what you're trying to do, use

var copy_initial_variable = initial_variable.slice(0);

Read Does JavaScript pass by reference? for more information.